Zivot - Lectures on Structural Change.pdf

Embed Size (px)

Citation preview

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    1/14

    Lectures on Structural Change

    Eric Zivot

    Department of Economics, University of Washington

    April 5, 2003

    1 Overview of Testing for and Estimating Struc-

    tural Change in Econometric Models

    1. Day 1: Tests of Parameter Constancy

    2. Day 2: Estimation of Models with Structural Change

    3. Day 3: Time Varying Parameter Models

    2 Some Preliminary Asymptotic Theory

    Reference: Stock, J.H. (1994) Unit Roots, Structural Breaks and Trends,in Handbook of Econometrics, Vol. IV.

    3 Tests of Parameter Constancy in Linear Mod-

    els

    3.1 Motivation

    Diagnostics for model adequacy

    Provide information about out-of-sample forecasting accuracy

    Within-sample parameter constancy is a necessary condition for super-exogeneity

    3.2 Example Data Sets

    3.2.1 Simulated Data

    Consider the linear regression model

    yt = + xt + t, t = 1, . . . , T = 200

    xt iid N(0, 1)t iid N(0, 2)

    1

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    2/14

    No structural change parameterization: = 0, = 1, = 0.5Structural change cases

    Break in intercept: = 1 for t > 100

    Break in slope: = 3 for t > 100

    Break in error variance: = 0.25 for t > 100

    Random walk in slope: = t = t1 + t, t iid N(0, 0.1) and 0 = 1.

    (show simulated data)

    3.2.2 US/DM Monthly Exchange rate data

    Let

    st = log of spot exchange rate in month t

    ft = log of forward exchange rate in month t

    The forward rate unbiased hypothesis is typically investigated using the so-calleddifferences regression

    st+1 = + (ft st) + t+1ft st = iUSt iDMt = forward discount

    If the forward rate ft is an unbiased forecast of the future spot rate st+1 thenwe should find

    = 0 and = 1

    The forward discount is often modeled as an AR(1) model

    ft st = + (ft1 st1) + utStatistical Issues

    st+1 is close to random walk with large variance

    ft st behaves like highly persistent AR(1) with small variance ft st appears to be unstable over time

    3.3 Chow Forecast Test

    Reference: Chow, G.C. (1960). Tests of Equality between Sets of Coeffi-cients in Two Linear Regressions, Econometrica, 52, 211-22.

    Consider the linear regression model with k variables

    yt = x0

    t + ut, ut (0, 2), t = 1, . . . , ny = X + u

    2

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    3/14

    Parameter constancy hypothesis

    H0 : is constant

    Intuition

    If parameters are constant then out-of-sample forecasts should be unbiased(forecast errors have mean zero)

    Test construction:

    Split sample into n1 > k and n2 = n n1 observations

    y =

    y1y2

    n1n2

    , X =

    X1X2

    n1n2

    Fit model using first n1

    observations

    1 = (X0

    1X1)1X01y1

    u1 = y1 X1121 = u

    0

    1u1/(n1 k)

    Use 1 and X2 to predict y2 using next n2 observations

    y2 = X21

    Compute out-of-sample prediction errors

    u2 = y2y2 = y2X21

    Under H0 : is constant

    u2 = u2 X2(1)

    and

    E[u2] = 0

    var(u2) = 2

    In2 + X2(X0

    1X1)1X02

    Further, If the errors u are Gaussian then

    u2 N(0,var(u2))u02var(u2)

    1u2 2(n2)(n1 k)21/2 2(n1 k)

    This motivates the Chow forecast test statistic

    ChowFCST(n2) =u02

    In2 + X2(X

    0

    1X1)1X02

    u2

    n221

    F(n2, n1 k)

    3

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    4/14

    Decision: Reject H0 at 5% level if

    ChowFCST(n2) > cv0.05

    Remarks:

    Test is a general specification test for unbiased forecasts

    Popular with LSE methodology

    Implementation requires a priori splitting of data into fit and forecastsamples

    3.3.1 Application: Simulated Data

    Chow Forecast Test

    n2Model 100 50 25No SC 1.121 1.189 1.331Mean shift 9.130*** 1.329* 1.061Slope shift 9.055*** 2.067*** 1.545*Var shift 0.568 0.726 0.864RW slope 2.183*** 1.302 0.550

    3.4 CUSUM and CUSUMSQ Tests

    Reference: Brown, R.L., J. Durbin and J.M. Evans (1975). Techniquesfor Testing the Constancy of Regression Relationships over Time, Journal ofthe Royal Statistical Society, Series B, 35, 149-192.

    3.4.1 Recursive least squares estimation

    The recursive least squares (RLS) estimates of are based on estimating

    yt = 0

    txt + t, t = 1, . . . , n

    by least squares recursively for t = k + 1, . . . , n giving nk least squares (RLS)estimates (k+1, . . . , T).

    RLS estimates may be efficiently computed using the Kalman Filter

    If is constant over time then t should quickly settle down near a com-mon value.

    4

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    5/14

    3.4.2 Recursive residuals

    Formal tests for structural stability of the regression coefficients may be com-puted from the standardized 1 step ahead recursive residuals

    wt =vt

    ft=

    yt 0

    t1xtft

    ft = 2h

    1 + x0t(X0

    tXt)1xt

    iIntuition:

    Ifi changes in the next period then the forecast error will not have meanzero

    wt are recursive Chow Forecast t-statistics with n2 = 1

    3.4.3 CUSUM statistic

    The CUSUM statistic of Brown, Durbin and Evans (1975) is

    CUSUMt =tX

    j=k+1

    wjw

    2w =1

    n knXt=1

    (wt w)2

    Under the null hypothesis that is constant, CUSUMt has mean zero andvariance that is proportional to t

    k

    1.

    3.4.4 CUSUMSQ statistic

    THE CUSUMSQ statistic is

    CUSUMSQt =

    Ptj=k+1 w

    2jPn

    j=k+1 w2j

    Under the null that is constant, CUSUMSQt behaves like a 2(t) and con-

    fidence bounds can be easily derived.

    3.4.5 Application: Simulated Data

    (insert graphs here)

    Remarks

    Ploberger and Kramer (1990) show the CUSUM test can be constructedwith OLS residuals instead of recursive residuals

    5

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    6/14

    CUSUM Test is essentially a test to detect instability in intercept alone

    CUSUM Test has power only in direction of the mean regressors

    CUSUMSQ has power for changing variance

    There are tests with better power

    3.4.6 Application: Exchange Rate Regression

    (insert graphs here)

    3.5 Nybloms Parameter Stability Test

    Reference:Nyblom, J. (1989). Testing for the Constancy of Parameters OverTime, Journal of the American Statistical Association, 84 (405), 223-230.

    Consider the linear regression model with k variables

    yt = x0

    t + t, t = 1, . . . , n

    The time varying parameter (TVP) alternative model assumes

    = t = t1 + t, it (0, 2i

    ), i = 1, . . . , k

    The hypotheses of interest are

    H0 : is constant 2i

    = 0 for all i

    H1 : 2i

    > 0 for some i

    Nyblom (1989) derives the locally best invariant test as the Lagrange multiplier

    test. The score assuming Gaussian errors isnXt=1

    xtt = 0

    t = yt x0t= (X0X)1X0y

    Define

    ft = xtt

    St =t

    Xj=1ft = cumulative sums

    V = n1X0X

    Note thatnXj=1

    ft = 0

    6

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    7/14

    Nyblom derives the LM statistic

    L = 1n2

    nXt=1

    StV1St

    =1

    n2tr

    "V1

    nXt=1

    StS0

    t

    #

    Under mild assumptions regarding the behavior of the regressors, the limitingdistribution of L under the null is a Camer-von Mises distribution:

    L Z10

    Bk()Bk()

    0d

    Bk

    () = Wk() Wk(1)W

    k() = k dimensional Brownian motion

    Decision: Reject H0 at 5% level if

    L > cv0.05

    Remarks:

    Distribution of L is non-standard and depends on k.

    Critical values are computed by simulation and are given in Nyblom,Hansen (1992) and Hansen (1997)

    Test is for constancy of all parameters

    Test is not informative about the date or type of structural change

    Test is applicable for models estimated by methods other than OLS

    Distribution of L is different if xt is non-stationary (unit root, determin-istic trend). See Hansen (1992).

    3.5.1 Application: Simulated Data

    Nyblom TestModel LcNo SC .332Mean shift 13.14

    Slope shift 14.13

    var shift .351RW slope 9.77

    7

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    8/14

    3.5.2 Application: Exchange rate regression

    Nyblom TestModel LcAR(1) 1.27

    Diff reg .413

    3.6 Hansens Parameter Stability Tests

    References

    1. Hansen, B.E. (1992). Testing for Parameter Instability in Linear Mod-els Journal of Policy Modeling, 14(4), 517-533.

    2. Hansen, B.E. (1992). Tests for Parameter Instability in Regressionswith I(1) Processes, Journal of Business and Economic Statistics, 10,321-336.

    Idea: Extension of Nybloms LM test to individual coefficients.Under the null of constant parameters, the score vector from the linear model

    with Gaussian errors is

    nXt=1

    xitt = 0, i = 1, . . . , kX(2t 2) = 0

    t = yt x0t

    2 = n1n

    Xt=1 2tDefine

    fit =

    xitt

    2t 2i = 1, . . . ki = k + 1

    Sit =tX

    j=1

    fij , i = 1, . . . , k + 1

    Note thatn

    Xi=1fit = 0, i = 1, . . . , k + 1

    3.6.1 Individual Coefficient Tests

    Hansens LM test for

    H0 : i is constant, i = 1, . . . , k

    8

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    9/14

    and forH0 :

    2 is constant

    is

    Li =1

    nVi

    nXt=1

    S2it, i = 1, . . . , k

    Vi =nXt=1

    f2it

    Under H0 : i is constant or H0 : 2 is constant

    Li Z10

    B1 ()B1 ()d

    Decision: Reject H0 at 5% level if

    Li > cv0.05 = 0.470

    3.6.2 Joint Test for All Coefficients

    For testing the joint hypothesis

    H0 : and 2 are constant

    define the (k + 1) 1 vectors

    ft = (f1t, . . . , f k+1,t)0

    St = (S1t, . . . , S k+1,t)0

    Hansens LM statistic for testing the constancy of all parameters is

    Lc =1

    n

    nXt=1

    S0tV1St =

    1

    ntr

    V1

    nXt=1

    StS0

    t

    !

    V =nXt=1

    ftf0

    t

    Under the null of no-structural change

    Lc

    Z1

    0

    Bk+1()Bk+1()d

    Decision: Reject H0 at 5% level if

    Lc > cv0.05

    Remarks

    9

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    10/14

    Tests are very easy to compute and are robust to heteroskedasticity

    Null distribution is non-standard and depends upon number of parameterstested for stability

    Individual tests are informative about the type of structural change

    Tests are not informative about the date of structural change

    Hansens L1 test for constancy of intercept is analogous to the CUSUMtest

    Hansens Lk+1 test for constancy of variance is analogous to CUSUMSQtest

    Hansens Lc test for constancy of all parameters is similar to Nybolomstest

    Distribution of tests is different if data are nonstationary (unit root, de-terministic trend) - see Hansen (1992), JBES.

    3.6.3 Application: Simulated Data

    Hansen TestsModel 2 JointNo SC .179 .134 .248 .503Mean shift 13.19 .234 .064 13.3

    Slope shift .588 5.11 .067 5.25

    var shift .226 .119 .376 .736RW slope .253 4.08 .196 4.4

    3.6.4 Application: Exchange rate regression contd

    Hansen TestsModel intercept slope variance JointAR(1) .382 .147 2.94 3.90

    Diff reg .104 .153 .186 .520

    4 Tests for Single Structural Change

    Consider the linear regression model with k variables

    yt = x0

    tt + t, t = 1 . . . , n

    No structural change null hypothesis

    H0 : t=

    10

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    11/14

    Single break date alternative hypothesis

    H1 : t= , t m = break datet = + , t > m and 6= 0k < m < n k =

    m

    n= break fraction

    Remarks:

    Under no break null = 0.

    Pure structural change model: all coefficients change (i 6= 0 for i =1, . . . , k)

    Partial structural change model: some coefficients change (i 6= 0 for somei)

    m = [ n], [] = integer part

    4.1 Chows Test with Known Break Date

    Assume: m or is knownFor a data interval [r , . . . , s] such that s r > k define

    r,s = OLS estimate of

    r,s = OLS residual vector

    SS Rr,s = 0

    r,sr,s = sum of squared residuals

    Chows breakpoint test for testing H0 vs. H1 with m known is

    Fn

    mn

    = Fn () =

    (SS R1,n (SS R1,m + SS Rm+1,n))/k(SS R1,m + SS Rm+1,n) /(n 2k)

    The Chow test may also be computed as the F-statistic for testing = 0 fromthe dummy variable regression

    yt = x0

    t + Dt(m)x0

    t+ t

    Dt(m) = 1 if t > m; 0 otherwise

    Under H0 : = 0 with m known

    Fn() F(k, n 2k)

    k Fn()d

    2(k)Decision: Reject H0 at 5% level if

    Fn() > F0.95(k, n k)k Fn() >

    20.95(k)

    11

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    12/14

    4.1.1 Application: Simulated Data

    Chow Breakpoint TestF200(0.5) F200(0.25) F200(0.75)No SC 0.808 0.081 1.55Mean shift 377*** 13.03*** 11.21***Slope shift 374*** 10.97*** 17.57***Var shift 1.071 0.117 1.204RW slope 80.14*** 4.058** 2.218

    4.2 Quandts LR Test with Unknown Break Date

    References:

    1. Quandt, R.E. (1960). Tests of Hypotheses that a Linear System Obeys

    Two Separate Regimes, Journal of the American Statistical Association,55, 324-330.

    2. Davies, R.A. (1977). Hypothesis Testing When a Nuisance Parameteris Present only Under the Alternative, Biometrika, 64, 247-254.

    3. Kim, H.-J., and D. Siegmund (1989). The Likelihood Ratio Test fora Change-Point in Simple Linear Regression, Biometrika, 76, 3, 409-23.

    4. Andrews, D.W.K. (1993). Tests for Parameter Instability and Struc-tural Change with Unknown Change Point, Econometrica, 59, 817-858.

    5. Hansen, B.E. (1997). Approximate Asymptotic P Values for Structural-Change Tests, Journal of Business and Economic Statistics, 15, 60-67.

    Assume: m or is unknown.Quandt considered the LR statistic for testing H0 : = 0 vs. H1 : 6= 0

    when m is unknown. This turns out to be the maximal Fn() statistic over arange of break dates m0, . . . , m1 :

    QLR = maxm[m0,m1]

    Fn

    mn

    = max

    [0,1]Fn()

    i =min

    = trimming parameters, i = 0, 1

    Remarks

    QLR is also know as Andrews supF statistic

    Trimming parameters 0 and 1 must be set

    Cannot have 0 = 1 and 1 = 1 because breaks are hard to identifynear beginning and end of sample

    Information about location of break can be used to specify 0 and1

    12

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    13/14

    Andrews recommends 0 = 0.15 and 1 = 0.85 if there is no knowl-edge of break date

    Implicitly, the break data m and break fraction are estimated using

    m = arg maxm

    Fn

    mn

    = m/n

    Under the null, m defined under the alternative is not identified. This isan example of the Davies problem.

    Davies (1977) showed that if estimated parameters are unidentified underthe null, standard 2 inference does not obtain.

    Under H0 : = 0, Kim and Siegmund (1989) showed

    k QLR sup[0,1]

    Bk()0Bk()

    (1 )Bk() = Wk() Wk(1) = Brownian Bridge

    Decision: Reject H0 at 5% level if

    k QLR > cv0.05

    Remarks

    Distribution of QLR is non-standard and depends on the number of vari-ables k and the trimming parameters 0 and 1

    Critical values for various values of 0 and 1 computed by simulation aregiven in Andrews (1993), and are larger than 2(k) critical values. For0 = 0.15 and 1 = 0.85

    5% critical valuesk 2(k) QLR k QLR1 3.84 8.8510 18.3 27.03

    P-values can be computed using techniques from Hansen (1997)

    Graphical plot of Fn() statistics is informative to locate the break date

    4.2.1 Application: Simulated data

    QLR or sup-F Test

    QLR

    bm

    b

    No SC 2.87 142 0.71Mean shift 377*** 101 0.51Slope shift 374*** 101 0.51Var shift 2.36 142 0.71RW slope 113*** 77 0.39

    (insert graphs here)

    13

  • 7/27/2019 Zivot - Lectures on Structural Change.pdf

    14/14

    4.2.2 Application: Exchange rate data

    QLR or sup-F TestQLR bm b

    AR(1) 12.13*** 1989:05 0.65Diff reg 4.08 1991:03 0.74

    4.3 Optimal Tests with Unknown Break Date

    References:

    1. Andrews, D.W.K. and W. Ploberger (1994). Optimal Tests Whena Nuisance Parameter Is Present Only Under the Alternative, Economet-rica, 62, 1383-1414.

    Andrews and Ploberger (1994) derive tests for structural change with anunknown break date with optimal power. These tests turn out to be weightedaverages of the Chow breakpoint statistics Fn(

    mn

    ) used to compute the QLRstatistic:

    ExpFn = ln

    1

    m2 m1 + 1m2X

    t=m1

    exp

    1

    2k Fn

    t

    n

    !

    AveFn =1

    m2 m1 + 1m2Xt=m1

    k Fn

    t

    n

    k = number of regressors being tested

    Remarks

    Asymptotic null distributions are non-standard and depend on k, 0 and1

    Critical values are given in Andrews and Ploberger; P-values can be com-puted using techniques of Hansen (1997)

    Tests can have higher power than QLR statistic

    Tests are not informative about location of break date

    4.4 Empirical Application

    Reference: Stock and Watson (199?), Journal of Business and Economic

    Statistics.

    14